Question

$\boxed{3x^2 - 4xy + 12x - 4y = 0}$

Answer 1

The solution to the equation is given by the relation: $y = \frac{3x^2 + 12x}{4(x + 1)}$

Answer 2

The equation $3x^2 - 4xy + 12x - 4y = 0$ is solved for $y$ by isolating $y$ terms: $-4xy - 4y = -3x^2 - 12x$ $-4y(x+1) = -3x^2 - 12x$ $y = \frac{-3x^2 - 12x}{-4(x+1)}$ $y = \frac{3x^2 + 12x}{4(x+1)}$ This relation defines the solution set for $(x, y)$, with the condition $x \neq -1$.